ARITHMETIC PROGRESSIONS IN CYCLES OF QUADRATIC POLYNOMIALS

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1 ARITHMETIC PROGRESSIONS IN CYCLES OF QUADRATIC POLYNOMIALS TIMO ERKAMA It is an open question whether n-cycles of complex quadratic polynomials can be contained in the field Q(i) of complex rational numbers for n 5. We prove that if such cycles exist, they cannot contain arithmetic progressions and that they are never superattracting. AMS 2000 Subject Classification: Primary 37F10; Secondary 11G30, 37F45. Key words: polynomial iteration, rational cycle, Mandelbrot set. 1. INTRODUCTION An n-cycle of a quadratic polynomial P is a set S of n complex numbers cyclically permuted by P. Thus, for each x S the numbers x 0 = x, x 1 = P (x 0 ), x 2 = P (x 1 ),..., x n 1 = P (x n 2 ) are distinct, and x 0 = P (x n 1 ). A complex rational n-cycle of P is an n-cycle contained in the field Q(i) of complex rational numbers. It is not uncommon that three points x i, x j and x k of an n-cycle of a quadratic polynomial form an arithmetic progression, so that x j x i = x k x j. This can occur for arbitrarily large values of n. However, for complex rational cycles the situation is much more special. We namely have Theorem 1. Let r, s and t be distinct points of a complex rational n- cycle of a quadratic polynomial P, and suppose that t s = s r. Then n = 3 and P is linearly conjugate to x 2 29/16. Theorem 1 solves a special case of the following Conjecture. Quadratic polynomials do not have complex rational n- cycles for any n 4. So far, this conjecture has been proved only for real rational four- and five-cycles (see [4] and [2]), and for complex four-cycles (see [1]). A nonrational n-cycle of a quadratic polynomial can contain arithmetic progressions for each n > 3. For example, let v be any root of the cubic equation 2v 3 v = 0. Then the polynomial x 2 + vx + v 1 has a REV. ROUMAINE MATH. PURES APPL., 54 (2009), 5 6,

2 442 Timo Erkama 2 4-cycle {1 v, 0, v 1, 2v(v 1)} which contains an arithmetic progression {1 v, 0, v 1}. However, four consecutive points of a cycle of a quadratic polynomial cannot form an arithmetic progression; this follows immediately from the Lagrange interpolation formula. To establish our results, it suffices to consider the special case of polynomials of the form P c (x) = x 2 + c where c is a complex constant. The dynamics of such polynomials can be studied by using a two-dimensional model introduced in [1]; here, we consider this model in homogeneous coordinates. In Section 3 we show that complex rational n-cycles of quadratic polynomials are never superattracting for n 3; it then follows that the centers of hyperbolic components of period n of the Mandelbrot set are not contained in Q(i) for n 3. We also prove that none of the points of a complex rational n-cycle of P c can be a Gaussian integer if n HOMOGENEOUS COORDINATES In [1] we showed that the dynamics of the family {P c } c C is equivalent to the dynamics of a single two-dimensional quadratic polynomial map F defined by (1) F (x, y) = (y, y 2 + y x 2 ) in the complex 2-space C 2. In particular, x is a periodic point of P c if and only if (x, P c (x)) is a periodic point of F with the same period. Thus, the map x (x, P c (x)) maps cycles of P c to cycles of F. In the study of rational cycles it is more convenient to consider this model in homogeneous coordinates. Let U = {(x, y, z) C 3 ; z 0}, and define G : U U and ψ : U C 2 by ( x G(x, y, z) = (yz, yz + y 2 x 2, z 2 ) and ψ(x, y, z) = z, y ) z for each (x, y, z) U. Two points of U are called equivalent if they have the same image under ψ. Such an image ψ(x, y, z) can be thought as the equivalence class of (x, y, z), and we shall denote it by [x, y, z]. Then F becomes the quotient map of G, so that the diagram U ψ C 2 G U ψ F C 2 is commutative. A periodic point of F has rational coordinates if and only if it is of the form [x, y, z] where x, y and z are integers. Similarly, points of a complex rational cycle of F are of the form [x, y, z] where x, y and z are

3 3 Arithmetic progressions in cycles of quadratic polynomials 443 Gaussian integers. If xy 0, we can then always choose x, y and z such that their greatest common divisor g.c.d. (x, y, z) in Z[i] is one. Theorem 2. Let [x, y, z] be a periodic point of F such that x, y and z 0 are Gaussian integers. Then there exists w Z[i] such that F ([x, y, z]) = [y, w, z]. If in addition xy 0 and g.c.d. (x, y, z) = 1, then g.c.d. (x, z) = g.c.d. (y, z) = 1. Proof. Denote by (x k, y k, z k ) the points of the orbit of (x, y, z) under G, so that (x, y, z) = (x 0, y 0, z 0 ) and G(x k, y k, z k ) = (x k+1, y k+1, z k+1 ) for each k. Note that here the sequences {x k }, {y k } and {z k } are in general not periodic. Let p be any prime in Z[i]. Lemma 1. If z 0 (mod p), then y 2 x 2 (mod p). Proof. We first prove that (2) x k 0 (mod p) and y k (y 2 x 2 ) 2k 1 (mod p). for each positive integer k. The case k = 1 is obvious because x 1 = yz and y 1 = yz + y 2 x 2. By induction, assume that (2) holds for some k. Since evidently z k = z 2k, we have x k+1 = y k z k 0 (mod p) and y k+1 = y k z k + yk 2 x2 k y2 k x2 k (mod p), and by the induction hypothesis y k+1 y 2 k (y2 x 2 ) 2k (mod p). Hence (2) holds for each k. If [x, y, z] has period n, we have [x n, y n, z n ] = [x, y, z]. In particular, y n z = z n y = z 2n y, so that y n = z 2n 1 y. Using (2), we conclude that and the assertion follows. (y 2 x 2 ) 2n 1 z 2n 1 y 0 (mod p), Corollary 1. [0, y, z] is periodic only if y = 0 or y = z. Proof. Suppose that y 0; we may then assume that g.c.d. (y, z) = 1. If p is a prime of Z[i] dividing z, by Lemma 1 y 2 0 (mod p). Since g.c.d. (y, z) = 1, we conclude that z is a unit of Z[i]. It follows that the orbit of [0, y, z] is integral, i.e., the coordinates of each point of the orbit in C 2 are Gaussian integers. In [1, Theorem 2] we proved that such an orbit must have period n 2. Now, n = 2, because y 0, and we conclude that F ([0, y, z]) = [yz, yz + y 2, z 2 ] = [y, 0, z]. This is possible only if y = z. Lemma 2. If z 0 (mod p ν ) for some ν 1, then y 2 x 2 (mod p ν ).

4 444 Timo Erkama 4 Proof. The case ν = 1 follows from Lemma 1. By induction, assume that the assertion holds for ν, and suppose that z 0 (mod p ν+1 ). Then there is µ Z[i] such that z = µp ν+1, and by the induction hypothesis y 2 x 2 = mp ν for some m Z[i]. Therefore, [x 1, y 1, z 1 ] = [yz, yz + mp ν, z 2 ] = [yµp ν+1, yµp ν+1 + mp ν, µ 2 p 2ν+2 ] = [yµp, yµp + m, µ 2 p ν+2 ]. Since [x 1, y 1, z 1 ] is also periodic, by Lemma 1, (yµp + m) 2 (yµp) 2 (mod p). It follows that m 2 0 (mod p), and we conclude that This completes the proof. y 2 x 2 = mp ν 0 (mod p ν+1 ). We can now prove Theorem 2. By Lemma 2, y 2 x 2 is divisible by all terms appearing in the prime factorization of z. Hence there exists m Z[i] such that y 2 x 2 = mz, and F [x, y, z] = [yz, yz + y 2 x 2, z 2 ] = [y, y + m, z]. Thus, we may choose w = y + m. To prove the last assertion, assume that g.c.d. (x, y, z) = 1. Let p be any prime in Z[i] dividing z. Then y 2 x 2 = mz implies that p divides either y x or y + x. This is possible only if p divides neither x nor y, because g.c.d. (x, y, z) = 1. Hence g.c.d. (x, z) = g.c.d. (y, z) = 1. As in [1], we say that a Gaussian integer is even if the sum of its real and imaginary parts is even; otherwise we call it odd. Obviously a nonzero Gaussian integer is even if and only if it is divisible by κ = 1 + i. Theorem 3. Suppose that [x, y, z] has period n 3 and g.c.d. (x, y, z) = 1. Then z is even and there is an n-periodic sequence {x k } of odd Gaussian integers such that x 0 = x, x 1 = y and (3) F ([x k, x k+1, z]) = [x k+1, x k+2, z] for each k 0. The sequence {x k } has the algebraic properties (4) (5) (x l+1 x k+1 )z = x 2 l x2 k for each l, k 0, n (x k + x k+ν ) = z n for each ν = 1,..., n 1. k=1 Moreover, the Gaussian integers x 0,..., x n 1 and z are pairwise relatively prime.

5 5 Arithmetic progressions in cycles of quadratic polynomials 445 Proof. The existence of a periodic sequence {x k } satisfying (3) follows from Theorem 2. To prove (4), we may assume that l = k + ν for some ν > 0. Then by (3) for each µ 0 we have [x µ+1, x µ+2, z] = F ([x µ, x µ+1, z]) = [x µ+1 z, x µ+1 z + x 2 µ+1 x 2 µ, z 2 ]. Comparison of the second components shows that x µ+2 z = x µ+1 z + x 2 µ+1 x2 µ and, therefore, (x µ+2 x µ+1 )z = x 2 µ+1 x2 µ. Addition of these equations for µ = k, k + 1,..., k + ν 1 proves (4). For l = k + ν it follows from (4) that (x k+ν+1 x k+1 )z = x 2 k+ν x2 k. Multiplication of these equations yields n 1 n 1 (x k+ν+1 x k+1 )z = (x 2 k+ν x2 k ). k=0 Since by periodicity x n+ν x n = x ν x 0, cancellation of the differences x k+ν x k proves (5). The rest of the proof depends on Lemma 3. Let λ be a Gaussian integer dividing x j x k such that g.c.d. (λ, z) = 1. Then λ divides x j+ν x k+ν for each ν = 1,..., n 1. Proof. The assertion follows immediately from (4) by induction on ν, for if λ divides x j+ν x k+ν, then it also divides x 2 j+ν x2 k+ν = (x j+ν+1 x k+ν+1 )z. Next, we show that z is even and that each x k is odd. Since n 3, at least two of the numbers x k are congruent mod κ. Then we can choose ν {1,..., n 1} such that the left-hand side of (5) is even. Hence z is even. Since by hypothesis g.c.d. (x, y, z) = 1, either x or y is odd while by (4) y 2 x 2 = (x 2 x 1 )z is even. This is possible only if both x and y are odd. Now using (4) for l = 0 we see that each x k is odd. It remains to prove that the Gaussian integers x 0,..., x n 1 and z are pairwise relatively prime. Application of Theorem 2 to [x 1, x 2, z] shows that g.c.d. (x 2, z) = 1, and continuation of the argument by induction proves that g.c.d. (x k, z) = 1 for each k. (Note that by Corollary 1 we need not worry about the possibility that some x k might be zero.) Finally, assume that x k and x k+ν have a common odd prime factor p for some ν > 0. From (4) we see that x k+ν+1 x k+1 is divisible by p 2. Lemma 3 then shows that x k+2ν x k+ν is divisible by p 2, so that x k+2ν is divisible by p. Now (4) implies that k=0 (x k+2ν+1 x k+ν+1 )z = (x k+2ν x k+ν )(x k+2ν + x k+ν ) is divisible by p 3. Similarly, x k+3ν+1 x k+2ν+1 will be divisible by p 4 etc. By periodicity, this leads to a contradiction because a fixed difference x k+ν x k

6 446 Timo Erkama 6 cannot be divisible by arbitrarily high powers of p. Thus x k+ν and x k are relatively prime, and the proof of Theorem 3 is complete. Corollary 2. Complex rational n-cycles of P c do not contain Gaussian integers for n 3. This is a refinement of Theorem 2 of [1], where we proved that a complex rational n-cycle of P c cannot consist of Gaussian integers for n 3. Proof. If [x, y, z] is a point of a complex rational n-cycle of P c with n 3, by [1, Theorem 2] the orbit of [x, y, z] is not integral, i.e., z > 1. Then x/z cannot be a Gaussian integer, because g.c.d. (x, z) = 1. Our next result shows that the numbers x k cannot have small prime factors if n is large. Theorem 4. Let I be a prime ideal of Z[i] containing one of the numbers x k of Theorem 3. Then the quotient ring Z[i]/I contains at least 2n 1 elements. Proof. Let π : Z[i] Z[i]/I denote the canonical projection, and let x 0,..., x n 1 and z be as in Theorem 3. We first prove that the restriction of π to the set S = {x 0,..., x n 1 } is one-to-one. Suppose that π(x i ) = π(x j ) for some i, j, and let p be a prime of Z[i] generating I. Then p divides x k, because x k I, so that g.c.d. (p, z) = 1 by Theorem 3. In particular, p is odd. Since π(x i ) = π(x j ), either x i = x j or p divides x i x j. Choose ν {0,..., n 1} such that x j+ν = x k. If p divides x i x j, by Lemma 3 it also divides x i+ν x j+ν = x i+ν x k, and we conclude that p divides x i+ν. This is possible only if x i+ν = x k, because the elements of S are relatively prime, and we conclude that x i+ν = x j+ν and, therefore, x i = x j. Since π is a homomorphism, the restriction of π to the set S = { x 0,..., x n 1 } is also one-to-one. Hence it remains to show that the intersection of the sets π(s) and π( S) consists of the point π(x k ) only. Indeed, if π(x i ) = π( x j ) for some i, j, we have π(x i + x j ) = 0, so that x i + x j is divisible by p. Since p z, by (5) this is possible only if x i = x j = x k. 3. SUPERATTRACTING CYCLES Let P (x) = αx 2 + βx + γ be an arbitrary quadratic polynomial where α, β and γ are complex and α 0. An n-cycle of P is superattracting if it contains a point x such that P (x) = 0. Theorem 5. Superattracting n-cycles of quadratic polynomials are not contained in Q(i) for n 3.

7 7 Arithmetic progressions in cycles of quadratic polynomials 447 Proof. Let ξ 0,..., ξ n 1 be points of an n-cycle of a quadratic polynomial P (x) = αx 2 +βx+γ with n 3, and suppose that ξ k Q(i) for 0 k n 1; we wish to prove that P (ξ k ) 0 for each k. We may of course assume that ξ k+1 = P (ξ k ) for 0 k n 1, by defining ξ n = ξ 0. For 0 k 2 these equations then form a nonsingular linear system with a unique solution {α, β, γ}. Since the coefficients of this system are contained in Q(i), the same must be true of the coefficients α, β and γ of P. The linear polynomial ζ(x) = αx + β/2 conjugates P to a normalized polynomial P c of the quadratic family, so that, (6) ζ P = P c ζ. In addition, the points x ν = ζ(ξ ν ) are distinct points of Q(i) for 0 ν n 1, and (6) shows that P c (x ν ) = x ν+1 for each ν. Hence the points x ν form an n-cycle of P c. Since n 3, by Corollary 1, x ν 0 for each ν. Differentiation of (6) then shows that P (ξ ν ) 0 for each ν. We conclude that the cycle {ξ ν } is not superattracting, and the proof is complete. The postcritical limit set of P c is the set of limit points of the sequence 0, P c (0), P c (P c (0)),... of iterated images of the critical point 0 of P c. The Mandelbrot set M consists of all c C such that the points of the above sequence form a bounded subset of C. An equivalent definition of M in terms of F was given in [1]. A component of the interior of M is called hyperbolic if for each point c of this component the postcritical limit set of P c is an n-cycle of P c for some n 1. It is conjectured that each interior component of M is hyperbolic but a complete proof has not yet been given [3]. Each hyperbolic component of M has a unique center c such that the postcritical limit set of P c is a superattracting n-cycle of P c for some n 1. Thus, the result below is a special case of Theorem 5. Corollary 3. Let c be the center of a hyperbolic component of the Mandelbrot set. Then c Q(i) only if the superattracting cycle of P c has period 2. We conclude that M has only two hyperbolic components with centers in Q(i); the corresponding values of c are 0 and PROOF OF THEOREM 1 Let P be as in Theorem 1. The linear conjugation in (6) maps complex rational cycles of P to complex rational cycles of P c and arithmetic progressions to arithmetic progressions. Thus we may assume that P = P c for some c.

8 448 Timo Erkama 8 Assume that r, s and t are distinct points of a complex rational n-cycle of P c forming an arithmetic progression as in Theorem 1, so that t s = s r. We have to prove that n = 3 and that P c (x) = x 2 29/16. Let [x, y, z] be an expression of (s, P c (s)) in homogeneous coordinates, so that (s, P c (s)) = ψ(x, y, z) and g.c.d. (x, y, z) = 1. Theorem 3 implies that there exists a periodic sequence of Gaussian integers x k such that x 0 = x, x 1 = y and F ([x k, x k+1, z]) = [x k+1, x k+2, z] for each k. Moreover, x k and z are relatively prime in Z[i] for each k. Let i and j be subscripts such that r = x i /z and t = x j /z. Then the Gaussian integers x i, x 0 and x j form an arithmetic progression, so that (7) x j x 0 = x 0 x i = m for some Gaussian integer m. Now, (5) shows that 2x 0 = x i + x j divides z n. However, since x 0 and z are relatively prime in Z[i], we conclude that x 0 must be a unit of Z[i]. Multiplying z and each x k by a unit, we may then obviously assume that x 0 = 1. Let Γ be the multiplicative semigroup of Gaussian integers generated by i and κ = 1 + i. Then Γ consists of nonzero Gaussian integers of the form uκ n, where u is a unit of Z[i] and n is a nonnegative integer. Note that the real and imaginary parts of elements of Γ are contained in Γ {0}. Lemma 4. Let p be an odd prime factor of z. Then p divides m and the numbers x 0 + x i = 2 m and x 0 + x j = 2 + m are in Γ. Proof. From (4) we see that p divides each number of the form x 2 l x2 k. Then p must divide either x l x k or x l + x k but not both, because x l and x k are relatively prime. If p m then, by (7), we conclude that p divides x j + x 0 = 2 + m as well as x 0 + x i = 2 m; so, p also divides the difference 2m, a contradiction. Thus p divides m. The last assertion of the lemma now follows, because the numbers x 0 x i and x 0 x j are divisible by m and, consequently, x 0 + x i and x 0 + x j cannot have common odd prime factors with z. Since m is even, it is divisible by κ. Denote by a and b the real and imaginary parts of m/κ, respectively. Then (x 0 + x i )/κ = 1 a i(1 + b) and (x j + x 0 )/κ = 1 + a i(1 b) are in Γ. It follows that the real and imaginary parts of 1 a i(1 + b) and 1 + a i(1 b) are contained in Γ {0}. The same is then true of their products 1 a 2 and 1 b 2. This is possible only if a as well as b is one of the numbers 0, ±1 and ±3. We conclude that either m Γ or m is of the form m = γd, where γ divides 2 and d is one of the three numbers 2 ± i and 3.

9 9 Arithmetic progressions in cycles of quadratic polynomials 449 Define X = 1 2 κ(x j 1 x i 1 ) and Y = 1 2 κ(x j 1 + x i 1 ). Since each x k is odd, X and Y are in Z[i], and (8) X 2 + Y 2 = i(x 2 j 1 + x 2 i 1). Since x j x 0 = x 0 x i = m, by (4) we have x 2 j 1 x2 n 1 = x2 n 1 x2 i 1. So, x 2 j 1 + x2 i 1 = 2x2 n 1 and, consequently, (9) X 2 + Y 2 = (X + iy )(X iy ) = 2ix 2 n 1. Because the Gaussian integers x k are relatively prime, X and Y cannot have common odd prime factors. The same must then be true of X+iY and X iy and, because their product is even, both must be even but not divisible by 2. It follows that there exists a unit u of Z[i], and relatively prime odd Gaussian integers α and β such that (10) X + iy = uκα 2 (11) X iy = 1 u κβ2. Then, by (9), x 2 n 1 = α2 β 2. Obviously, we may choose α and β such that (12) x n 1 = αβ. Since u 4 = 1, by squaring (10) and (11) and subtracting we find that (13) 4iXY = iκ 2 (x 2 j 1 x 2 i 1) = u 2 κ 2 (α 4 β 4 ). Here, by (4), x 2 j 1 x2 i 1 = (x j x i )z = 2mz, and we conclude that (14) α 4 β 4 = κ 2 u 2 mz. It remains to solve this equation for possible values of m and z. purpose we can use the result below which was proved in [1]. and For this Lemma 5. Let α, β, γ and δ be Gaussian integers such that γ divides 2 (15) α 4 β 4 = γδ 2. Then αβδ = 0. This lemma implies easily that z Γ. In fact, we have seen earlier that either m Γ or m is of the form m = γd, where γ divides 2 and d is one of the three numbers 2 ± i and 3. On the other hand, because z divides x 2 j x2 i = 2m(x j + x i ) = 4m, either z Γ or mz is of the form ρd 2 for some ρ Γ. In the latter case the right-hand side of (14) has the same form as the right-hand side of (15), in contradiction with Lemma 5. Hence z Γ. If m Γ as well, the right-hand sides of (14) and (15) have again the same form, contradicting the fact that αβmz 0. Thus, it remains to study the case when z Γ and m = γd for some γ dividing 2.

10 450 Timo Erkama 10 In this case d divides one, and only one of the four factors α±β and α±iβ of the left-hand side of (14), while the remaining three factors are in Γ. At least two of these three factors must have modulus 2, because if e.g. α + β and α iβ were both divisible by κ 3, then α = 1 κ [i(α + β) + (α iβ)] would be even. We conclude that α as well as β has modulus 2 2, so that the only possible prime factors of α and β are 2+i and 2 i. Since α 4 β 4 0, at least one of the numbers α 4 and β 4 cannot be a unit and is therefore contained in a prime ideal I generated by either 2 + i or 2 i. But then x n 1 = αβ also is contained in I. By Theorem 4, the quotient ring Z[i]/I contains at least 2n 1 elements. Hence 2n 1 5, and we conclude that n 3. It remains to show that c = 29/16. By interchanging i and j if necessary, we may assume that P (r) = s, P (s) = t and P (t) = r. It then follows from the Newton interpolation formula that P (x) = s + (x r) 3z (x r)(x s) 2m = [ 3z 2 x 2 + (6 m)zx + 2m 2 + 3m 3]/2mz. The normalization of P = P c now implies that 3z 2 = 2mz and 6 m = 0. Then m = 6 and z = 4, so that c = (2m 2 + 3m 3)/2mz = 29/16. This completes the proof of Theorem 1. If r, s and t form an arithmetic progression as in Theorem 1, they satisfy a linear relation r 2s + t = 0. There are many similar relations which cannot be satisfied by points of complex rational n-cycles of the quadratic family for n 4. For example, the relation 3r + 4s 2t = 0 is impossible by Theorem 3, because each x k is odd. A less obvious example is given by the relation r + 2s t = 0, whose study is left for the interested reader. REFERENCES [1] Timo Erkama, Periodic orbits of quadratic polynomials. Bull. London Math. Soc. 38 (2006), 5, [2] E.V. Flynn, Bjorn Poonen and Edward F. Schaefer, Cycles of quadratic polynomials and rational points on a genus-2 curve. Duke Math. J. 90 (1997), 3, [3] Curtis T. McMullen, Frontiers in complex dynamics. Bull. Amer. Math. Soc. (N.S.) 31 (1994), 2, [4] Patrick Morton, Arithmetic properties of periodic points of quadratic maps. II. Acta Arith. 87 (1998), 2, Received 17 December 2008 University of Joensuu FI Joensuu Finland timo.erkama@joensuu.fi